# Polynomial Graphs Questions with Solutions

How to use the properties of the polynomial graphs to identify polynomials. Grade 12 math questions with detailed solutions and graphical interpretations are presented.

## Questions on Graphs of Polynomials

1. Question 1
Give four different reasons why the graph below cannot possibly be the graph of the polynomial function $p(x) = x^4-x^2+1$. .

Solution
The four reasons are:
1) The given polynomial function is even and therefore its graph must be symmetric with respect to the y axis. The given graph is not symmetric with respect to the y axis.
2) The given polynomial function does not have real zeros (discriminant = -3 : negative). The given graph has x intercepts which must corresponds to real zeros.
3) The y intercept calculated using p(x)( p(0) = 04 - 02 + 1 = 1) is positive. The y intercept of the graph is negative.
4) Having a leading coefficient ( = 1) positive and an even degree ( = 4), the polynomial must have a graph with the right and left both rising. In the given graph, they are both falling.

2. Question 2
Match the polynomial functions to their graphs where all x intercepts are shown.
$f(x) = (x+1)(x-1)^2(x+2)^2$ $g(x) = -(x+1)(x-1)^4$ $h(x) = (x+1)(x-1)^3(x-3)$ $i(x) = (x+1)^2(x-2)^3$ $j(x) = (x+1)^2(1-x)(x-2)^2$ $k(x) =-(x+1)^2(x-1)^2(x-3)$ .

Solution
According to their equations, all 6 given polynomial functions are of degree 5. However their leading coefficients are of different signs. We classify the 6 polynomials into 2 groups: I and II
Group I - Given polynomials with positive leading coefficients
f(x) = (x + 1)(x - 1)2(x + 2)2
h(x) = (x + 1)(x - 1)3(x - 3)
i(x) = (x + 1)2(x - 2)3
Having degree 5 (odd) and leading coefficients positive, each of the graphs of the above polynomials (f, h and i) has the following graphical properties:
as   x ____> ∞   ,   y ____> ∞   (the right hand side of the graphs rises)
as   x ____> - ∞   y ,   ____> - ∞   (the left hand side of the graph falls)
The given graphs in parts a) c) and e) have the above properties with different x intercepts and their multiplicities. Hence
Polynomial f(x) = (x + 1)(x - 1)2(x + 2)2 has a zero of multiplicity 1 at x = -1 , a zero of multiplicity 2 at x = 1 and a zero of multiplicity 2 at x = - 2 and should correspond to the graph in part e).
Polynomial h(x) = (x + 1)(x - 1)3(x - 3) has a zero of multiplicity 1 at x = -1 , a zero of multiplicity 3 at x = 1 and a zero of multiplicity 1 at x = 3 and should correspond to the graph in part a).
Polynomial i(x) = (x + 1)2(x - 2)3 has a zero of multiplicity 2 at x = -1 and a zero of multiplicity 3 at x = 2 and should correspond to the graph in part c).
Group II - Given polynomials with negative leading coefficients
The polynomial functions g, j and k, when expanded, have leading coefficients that are negative.
g(x) = - (x + 1)(x - 1)4
j(x) = (x + 1)2(1 - x)(x - 2)2
k(x) = - (x + 1)2(x - 1)2(x - 3)
Having degree 5 (odd) and leading coefficients negative, each of the graphs of the above polynomials (g, j and k) has the following graphical properties:
as   x ____> ∞   ,   y ____> - ∞   (the right hand side of the graphs falls)
as   x ____> - ∞   y ,   ____> ∞   (the left hand side of the graph rises)
The given graphs in parts b) d) and f) have the above properties with different x intercepts and their multiplicities. Hence
Polynomial g(x) = - (x + 1)(x - 1)4 has a zero of multiplicity 1 at x = -1 , a zero of multiplicity 4 at x = 1 and should correspond to the graph in part f).
Polynomial j(x) = (x + 1)2(1 - x)(x - 2)2 has a zero of multiplicity 2 at x = -1 , a zero of multiplicity 1 at x = 1 and a zero of multiplicity 2 at x = 2 and should correspond to the graph in part d).
Polynomial k(x) = - (x + 1)2(x - 1)2(x - 3) has a zero of multiplicity 2 at x = -1 , a zero of multiplicity 2 at x = 1 and a zero of multiplicity 1 at x = 3 and should correspond to the graph in part b).